Problem: The equation of hyperbola $H$ is $\dfrac {(x+4)^{2}}{64}-\dfrac {(y+3)^{2}}{16} = 1$. What are the asymptotes?
Explanation: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y+3)^{2}}{16} = - 1 + \dfrac {(x+4)^{2}}{64}$ Multiply both sides of the equation by $16$ $(y+3)^{2} = { - 16 + \dfrac{ (x+4)^{2} \cdot 16 }{64}}$ Take the square root of both sides. $\sqrt{(y+3)^{2}} = \pm \sqrt { - 16 + \dfrac{ (x+4)^{2} \cdot 16 }{64}}$ $ y + 3 = \pm \sqrt { - 16 + \dfrac{ (x+4)^{2} \cdot 16 }{64}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y + 3 \approx \pm \sqrt {\dfrac{ (x+4)^{2} \cdot 16 }{64}}$ $y + 3 \approx \pm \left(\dfrac{4 \cdot (x + 4)}{8}\right)$ Subtract $3$ from both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{1}{2}(x + 4) -3$